Question

The number of ants per acre in the forest is normally distributed with mean 44,000 and standard deviation 12,288. Let X = number of ants in a randomly selected acre of the forest. Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(,)

b. Find the probability that a randomly selected acre in the forest has fewer than 45,572 ants.

c. Find the probability that a randomly selected acre has between 45,750 and 58,877 ants.

d. Find the first quartile.  ants (round your answer to a whole number)

Answer 1

Solution :

(a)

X ~ N(44000 , 122882)

(b)

P(x < 45572) = P[(x - ) / < (45572 - 44000) / 12288]

= P(z < 0.13)

= 0.5517

Probability = 0.5517

(c)

P(45750 < x < 58877)

= P[(45750 - 44000) / 12288) < (x - ) /  < (58877 - 44000) / 12288) ]

= P(0.14 < z < 1.21)

= P(z < 1.21) - P(z < 0.14)

= 0.3312

Probability = 0.3312

(d)

P(Z < -0.674) = 0.25

z = -0.674

Using z-score formula,

x = z * +

x = -0.674 * 12288 + 44000 = 35718

First quartile = 35718